Definition 3.1.
A function
is said to be a solution of (1.1) if
satisfies (1.1).
In what follows one assumes that
One needs the following auxiliary result.
Lemma 3.2.
. Let
. Then the function defined by
is the unique solution of the boundary value problem
where
Proof.
Let
be a solution of the problem (3.2). Then integratingly, we obtain
Hence
where
Now, multiply (3.6) by
and integrate over
, to get
Thus,
Substituting in (3.6) we have
Therefore
Set
Note that
Our first result reads
Theorem 3.3.
Assume that
is an
-Carathéodory function and the following hypothesis
(A1) There exists
such that
holds. If
then the BVP (1.1) has a unique solution.
Proof.
Transform problem (1.1) into a fixed-point problem. Consider the operator
defined by
We will show that
is a contraction. Indeed, consider
Then we have for each 
Therefore
showing that,
is a contraction and hence it has a unique fixed point which is a solution to (1.1). The proof is completed.
We now present an existence result for problem (1.1).
Theorem 3.4.
Suppose that hypotheses
(H1) The function
is an
-Carathéodory,
(H2) There exist functions
and
such that
are satisfied. Then the BVP (1.1) has at least one solution. Moreover the solution set
is compact.
Proof.
Transform the BVP (1.1) into a fixed-point problem. Consider the operator
as defined in Theorem 3.3. We will show that
satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof will be given in several steps.
Step 1 (
is continuous).
Let
be a sequence such that
in
Then
Since
is
-Carathéodory and
then
Hence
Step 2 (
maps bounded sets into bounded sets in
).
Indeed, it is enough to show that there exists a positive constant
such that for each
one has
.
Let
. Then for each
, we have
By (H2) we have for each 
Then for each
we have
Step 3 (
maps bounded set into equicontinuous sets of
).
Let
,
and
be a bounded set of
as in Step 2. Let
and
we have
As
the right-hand side of the above inequality tends to zero. Then
is equicontinuous. As a consequence of Steps 1 to 3 together with the Arzela-Ascoli theorem we can conclude that
is completely continuous.
Step 4 (A priori bounds on solutions).
Let
for some
. This implies by
that for each
we have
Then
If
we have
Thus
Hence
Set
and consider the operator
From the choice of
, there is no
such that
for some
As a consequence of the nonlinear alternative of Leray-Schauder type [15], we deduce that
has a fixed point
in
which is a solution of the problem (1.1).
Now, prove that
is compact. Let
be a sequence in
, then
As in Steps 3 and 4 we can easily prove that there exists
such that
and the set
is equicontinuous in
hence by Arzela-Ascoli theorem we can conclude that there exists a subsequence of
converging to
in
Using that fast that
is an
-Carathédory we can prove that
Thus
is compact.